0387986995Basic TheoryC
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I. FUNDAMENTAL THEOREMS OF ODES |
Problem 1-3-6. Show that for any fixed t > 0, there exists a real-valued function 0(x) such that
(1)0(x) is continuous on an interval 0 < C-x <- 6 for some small positive number 6 which depends on (,
(2)fix) is continuous on an interval 0 < - z < 6,
(3)the function
1 + (: - x)4(x)
y(x) -
F( - x)
satisfies the differential equation
(Eq) |
dxy - 2xyL |
y2 |
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y3 |
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ono<x<6.
Answer.
Step 1. Ifwesett=£-randy = , the given differential equation is changed
to
t
(1)t2u" - 2tu' + 2u + 9(tu' - u)u = 2t(tu' - u)u + tsu-3 - tut,
where f = dt
Step2. If wesetu= 1+u'1 and tu' |
differential equation (1) becomes |
=w2,
_ - 2(1 + wl) + 3w2 - 2(w2 - (1 + wl))(1 + w1)
(2)
+ 2r1t(w2 - (1 +wl))(1 + w1)
+ t6 (1 + w1)-3 - |
(1 + w1)2. |
If we set t = 0, wl = 0, and w2 = 0, the right-hand side of (2) vanishes. Now, (2) can be written in the form
td!W--' = W2, dt
t dw2 =2w1+w2-2w1(w2-wl)
(3) |
dt |
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+ 2{-lt(w2 - (I + w1))(1 + w1) |
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+ t6t4(1 + w1)-3 - tt-1(I + wl)2. |
3. SOME GLOBAL PROPERTIES OF SOLUTIONS |
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Step 3. Write (3) in the following form:
(4) |
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te = f2w +'Pp) + tG(t,w"), |
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where |
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_ wl |
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1 |
0 |
w- [w2]' |
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['2 1] |
F(w) -[2w1(w2-wi)J' |
with
G(t,w) = 22-'(w2 - (1 +w1))(1 +w1)
(5)
+ t5e(1 + w1)-3 - t:-'(1 + wl)2.
Two eigenvalues of f2 are -1 and 2, which are distinct. Let us diagonalize SZ by the transformation
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w" = Pv", |
where P = |
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Then, (4) becomes |
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(6) |
tv = Av + P-' F(PV) + tP-1G(t, Pv), |
A = [ O1 } . |
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Set ii= ti. Then, t9 = ti+t2z and F(Pv-) = t2F(Pz). Therefore, (6) becomes
(7) ti' = Aoz + At' zl,
where Ao = [_2 0 I and f (t, z) = tP-'F(Pz) + P-'C(t, tPa). If t > 0 and
if tl1 and tll are small, there exist two positive constants Ko and L such that
If(t, z-) <- tLli - {I and If(t,z)1 <- Ko + tL17.
Step 4. We change (7) to the following system of integral equations:
zi(t) = t'2 fot fi(r,z(t))dt
(8)
z2(t) = d + t ft r 2f2(t, t(t))dt, o
where to 1is a sufficiently small positive number and c is an arbitrary constant, and
z' = (21 and f = 2 |
J . We want to construct a bounded solution of (8) on |
J |
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I. FUNDAMENTAL THEOREMS OF ODES |
0 < t < to. To do this, first note that, if IZ(t)I < K on 0 < t < to, we obtain
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Izt(t)I < t-2 fo rI f1(r, zlr))Idr < t-2 t r(Ko + rLK)dr |
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0 |
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20+3LK < 20+ 3LK, |
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to |
(9) |
Izs(t)I < Iclt + t f r-21 f2(r, z(r))Idr |
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f |
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t |
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to |
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< Icit + t |
r-2(Ko + rLK )dr |
t
= Iclt+t (' - 1 )Ko+t (In (12)) LK
< Iclto + K0 + e-1 t0LK.
For a positive number co, fix another positive number K so that co + Ko < 2 and
then fix t0 > 0 so that to < 1, toL < 1, and toKIPI < 1. Here, IPI = max{IP.kl
1 < j, k < 2}, where P = (Pjkj? k=1. Define successive approximations by
t_2Jo |
m > 0- |
i (t, C) = |
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d +t JJJ( |
r-2f2(r,-im(r,c))dr |
to
It can be shown that the successive approximations converges uniformly for Icl < Co and 0 < t < to. This establishes the existence of a bounded solution of the integral equation (8). Thus, we can complete the construction of solution (Sol) of equation
(Eq).
1-4. Analytic differential equations
So far, it has been assumed that the independent variable t and the unknown quantity y are real. Since a complex number is a two-dimensional real vector, every result in the previous sections can be extended to the case when the entries of g' and of the function f (t, g) are complex. Furthermore, if the function f is analytic with respect to (t, yam), we can construct analytic solutions by means of successive approximations. Denote by C the set of all complex numbers. The set of all n-
column vectors with complex entries is denoted by C".
4. ANALYTIC DIFFERENTIAL EQUATIONS |
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Theorem 1-4-1. Assume that each entry of a Cn-valued function f (z, yam) is given by a power series in (z, yl, y2, ... , yn) which converges in a polydisk:
D= ((z, y-) : Iz - zol < a, ly - col < b},
where z, yl, y2, , y" are complex quantities. Assume also that there exist positive numbers M and L such that
If(z,-)I < /M |
in |
D, |
(z,y,) E D (j = 1,2). |
l 0z, 91) - 1(z, g2)1 5 Lull - y21 |
whenever |
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Define successive approximations by |
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1m0(z) = 4o, |
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om(Z)=co+J f((,0-m-1( ))d(, |
m>1. |
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0 |
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where the path of integration may be taken to |
1be the line segment zaz joining zo to |
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z in the complex z-plane. Set a = min I a, M }. Then,
(i)For every m, the entries of the C"-valued))) function 4m(z) are power series in z - zo which are convergent in the open disk A = {z : iz - zoj < a},
(ii)the sequence { ,,,(z) m = 0,1, ... } satisfies the estimates
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m |
on 0, m = 0, 1, ... , |
,4n+l(Z) - om(Z)I |
IZ - zojm+l |
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II ,c |
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(nl + 1),
(iii) the sequence {¢m(z) m = 0, 1,... } converges to
¢(z) = o(z) + |
(mk(z) - |
(Z)) |
k=1
uniformly in 0 as m -+ +oo,
(iv)the entries of the C"-valued function b(z) are power series in z - zo which are convergent in 0,
(v) the function Q(z) is the unique solution to the initial-value problem
d = f(z,y), y(zo)
The proof of this theorem is left to the reader as an exercise. The reader must notice that even if a sequence of analytic functions fn(t) of a real variable t converges to a function f (t) uniformly on an interval Z, the limit function f (t) may not be analytic at all. For example, any continuous function can be uniformly approximated by polynomials on a bounded closed interval. In order to derive analyticity of f (t), uniform convergence must be proved on a domain in the complex t-plane.
Observation 1-4-2. The estimates
MLm |
IZ - ZOlm+1 in A, |
n1 = 0, 1,... |
m+1(Z) - & (Z) 1 :5 (m+ 1)I |
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I. FUNDAMENTAL THEOREMS OF ODES |
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+00 |
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imply that there exists a power series Eck(z - zo)k such that |
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k=0 |
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M |
m=0,1,2,.... |
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&(z) = E ck(z - z,)' +O(Iz - zol"'+1), |
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k=O
Observation 1-4-3. Since
m
&(Z) = $O(Z) + E (&(Z)
k=1
00
we obtain (z) - ¢,,,(z) = O(Iz - zol'"+1) and, hence, (z) _ >2Ck(Z - ZO)k.
Observation 1-4-4. Set |
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k=0 |
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(Z) = |
Ck(Z - ZO)k + 1: Cm.k(Z - Zo)k. |
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k=0 |
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k=m+1 |
Since |
cb1 < b in the disk 0, it follows that |
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1Cm,kI |
a |
(k = m + 1, m + 2,...). |
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Note that |
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4.,k = |
1 |
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0(z) - ca |
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2?ri |
Izo1=P (z - zo) k+1 |
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for any positive number p < a. This, in turn, implies that |
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,n |
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oo |
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Ck(Z-ZO)kl :5 b |
IZ aZOlk |
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k=O |
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k=m+1 |
=blzzD 1
I
a 1-IZ
ZIDI
in the disk A.
Example 1-4-5. In the case of the initial-value problem dzy = (1 - z)y2, y(0) = 2,
we obtain
2
O(z) = 2 E(k + 1)zk = (1 - Z)2
k=O
EXERCISES I |
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by using the successive approximations
Oo(z) = 2 ,
01(z) = 2 + 4z - 2z2,
02(x) = 2 + 4z + 6x2 - 83 - 6x4 + 4x5 |
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2x4 |
2825 |
82zs |
¢3(z) = 2 + 4z + 6z2 + 8x3 _ |
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256z7 |
124z8 |
68z9 |
46z10 |
56z" |
+ 63 + |
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+ 9 |
22z12 |
4z13 |
2z14 |
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9 + 9 - 63
Combining Theorem 1-3-5 and Theorem 1-4-1, we obtain the following theorem:
Theorem I-4-6. If all the entries of an n x n matrix A(t) and C"-valued function b(t) are analytic on an interval Z, then every solution of linear differential equation (1.3.1) is analytic on Z.
EXERCISES I
I-1. Solve the initial-value problem
d22 + 2y L = 0, |
y(r) = ,lo, b (r) = 171 |
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The reader must consider various cases concerning the initial data (r, i |
, m)1). |
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I-2 Show that the function f (x y) _ |
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satisfies the |
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(3 - (x - 1)2)(9 - (y - 5)2) |
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Lipschitz condition
If(x,yl)-f(x,y2)I < IY1Y2I
if
Ix-1I 5 f, Iyi-5I < 2, lye-51 < 2.
1-3. Show that the initial-value problem
(P) |
dy _ |
sm |
x3 + 3x + 1 |
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dx - |
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101-y2) |
y(5) = 3,
has one and only one solution on the interval Ix - 51 < 7.
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I. FUNDAMENTAL THEOREMS OF ODES |
1-4. Suppose that u(x) is continuous and satisfies the integral equation
x
u(x) = J sin(u(t))u(t)pdt
0
on the interval 0 < x < 1. Show that u(x) = 0 on this interval if p > 0. What would happen if p < 0?
u-p
Hint. In case p < 0, the problem is reduced to the initial-value problem sin(u) du = dx, u(O) = 0. Thus, the solution u(x) is not identically zero and has a branch point at x = 0.
1-5. Show that if real-valued continuous functions f (x), g(x), and h(x) satisfy the inequalities
f(x) ? 0, |
g(x) < h(x) + |
f |
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J0*'C |
on an interval 0 < x < xo, then |
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g(x) < h(x) + f |
= |
exp |
[JZ |
f(n)dn, } di |
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ono < x < xo. |
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Hint. If we put y(x) = J f |
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then LY < f(x)h(x) + f (x)y and y(O) = 0. |
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1-6. Let a(t) be a real-valued function which is continuous on the interval I = It:
0 < t < 11. For every positive integer rn, set
1, 2, ... } converges uniformly on the interval Z. Also, find |
lim f,,,. |
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fm (t) = I + ta(0) |
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m-oo |
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for 0 < t < 1 |
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-m |
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and |
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fm(t)= L=0(1+ma(m))]( l+lt-m/a\m)j |
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fork |
< t < k + 1 and k = 1, 2, ... , rn - 1. Show that the sequence {fm :M= |
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m |
m |
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Hint. For large m, we have 1 + m a (m / |
> 0 and |
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exp [ma(m/ - m21 < 1 + ma \m! < exp |
[M1 |
h |
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a(m)] . |
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for some positive number K. Also
fta(r)dr-Lmn(mj-(t-mja(ml)`
E,,,
M -0 |
Em = |
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for some positive e,,, such that m-» too |
0. |
EXERCISES I |
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1-7. Let f (t, y) be a real-valued function of two independent variables t and y which is continuous on a region:
D = {(t, y) : 91(t) 5 y <- 92(t), r1 < t < r2}.
Suppose that
(i) 91, g'1, 92, and g2 are continuous on rl < t < T2,
(ii) 91(t) < 92(t) on rl < t < T2,
(iii)
gi(t) < f(t,9i(t)), |
rl < t < rz, |
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f 9t> f(t, 92(t), |
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z |
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(iv) Tl < t0 < r2 and g1(to) < CO < 92(to)-
Show that there exists a function 0(t) such that
(1)4 and 0' are continuous on to < t < T2 and (t,4i(t)) E D on to < t
(2)Q'(t) = f (t, O(t)) on to < t < 72 and 0(to) = co.
1-8. Set
0 |
for |
f(u) = { f |
for |
- oo < y 5 0,
0 < y < +oo.
Also, let rt(x) be an arbitrary continuous function of x on an interval 0 < x < xo, where xo is a positive number. Define successive approximations by
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rt(r) |
for |
in = 0, |
tim(x) = |
f (ym_l(t))dt |
for |
in = 1, 2, |
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Show that lim ym(x) exists uniformly on the interval 0 < x < xo. m-+oo
Hint. For this problem, see [Na3, pp. 143-160; in particular Bemerkung 2].
1-9. Let B be a Banach space over the field R of real numbers. Also, let f (t, x) be a B-valued function of the variables (t, x) E R x B. Consider an initial-value problem
(IP) |
d = f (t, x), |
X(r) _ . |
where (r, {) is a given point in R x B. Assume that f satisfies the following condi-
tions:
(i) f(t,x)iscontinuous onaregion R={(t,x)ERxB:It-rl <a, IIx-t:ll<b},
where a and b are positive numbers and II II is the norm of B,
(ii) f(t.x) also satisfies the Lipschitz condition IIf(t,xl) -f(t,x2)II 5 LIIX1 -X211 whenever (t, xl) E R and (t, x2) E R, where L is a positive constant.
Show that
(1) there exists a positive number Af such that Ilf(t,x)II < M for (t,x) E R,
(2) problem (IP) has one and only one solution on the interval It - rI < or, where |
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a= min |
b |
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a, |
I. |
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b
Hint. See [Huk7].
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1. FUNDAMENTAL THEOREMS OF ODES |
I-10. Let B be a Banach space over the field R of real numbers and let C be a nonempty compact subset of B. Also, on a bounded interval Z, let F be an infinite and equicontinuous set of C-valued functions. Show that F contains an infinite sequence which is uniformly convergent on Z.
I-11. Let B be a Banach space over the field R of real numbers and let C be a nonempty compact and convex subset of B. Assume that a C-valued function f (t, x) is continuous on a region
R={(t,x)ERxB:It-rl <-a,llx-t1l <-b},
where r E R, t E B, a > 0, b > 0, and 11 11 is the norm of B. Show that
(1) there exists a positive number M such that IIf(t,z)115 M for (t,x) E R,
(2) the initial-value problem - = f (t, x), x(r) |
has a solution on the interval |
it - rl < or, where a = min ( a, f ). |
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Hint. Use a method similar to that of Remark 1-2-7. See, also, [Huk7l.
1-12. Let f (t, x") be Re-valued and continuous for (t, x") E Rr+t. Assume also that
E is a nonempty closed set in R"+t. Show that in order that for every (r,)) E E, there exists a real number a(r, depending on (r, ) such that r < a(-r, {) and
that the intial-value problem di = f (t, xF), x(r) _ has a solution s = ¢(t, r,
satisfying the condition (t, ¢(t, r, ) E E on an interval r < t < o(r), it is necessary and sufficient that for every (r, f E E and every positive number e, there exists a point (a, S) E E such that
0<a-r<e and |
< E, |
where (a, ) generaly depends on (r, {, e).
Hint. The necessity is easy to prove. For sufficiency, use an idea similar to that of
Remark 1-2-7. See, also, [Huk6].
1-13. Assume that f (x, yt, y2,... , y,,) is real-valued, continuous, and continuously
differentiable function of (x, yt, ... , |
on a region R. = {(x, yl,... , |
a < x < |
b, lyj I < r (j = 1,... , n)), where a and b are real numbers such that a < b, and r is
a positive number. Assume, also, that (1) c is a real number such that a < c < b,
l
(2) 6(x) is a real-valued solution of !° = f (x, y, Ly, ... , fn-Y) on an interval
lz - cl < p, where p is a positive number such that a < c - p < c + p < b, (3)
r (k = 0,1,... n - 1). and (4) f(x,0,... ,0) = 0 on a < x < b.
dx*
Show that if there eixsts a sequence {ch : h = 1, 2, ... } of real numbers such that
Um ch = c and that ¢(ch) = 0 (h = 1, 2, ... ), then O(x) = 0 identically on h_+00
a<x<b.
EXERCISES I |
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1-14. Let u(x, y) be a continuously differentiable function of two independent variables (x, y) in a domain D = {(x, y) : y > 0, a + Ay < x < b - Ay}, where A is a positive number and the two quantities a and b are real numbers such that a < b.
Assume that u(x, y) is continuous on the closure f) of D and satisfies the condition
< AI e I +BluF+C
on D, where B and C are positive numbers. Set M = max(Iu(x,0)(: a < r < b}.
Show that
u(x, y) 1 5 Me 11 + (eBV - 1)
on D.
B
Hint. The last inequality is the Haar inequality. (See [Haa1, [Na5, pp. 51-561, and
[Har2, pp. 140-141]).
1-15. Let H(u, t, x, q) be a function defined on an open set in R4 containing the point (u, t, x, q) = (0, 0, 0, 0) and satisfy a uniform Lipschitz condition with respect to (u, q). Let O(x) be a function of class C' satisfying 0(0) = 0 and 0=(0) = 0.
Show that the initial-value problem
ut + H(u, t, x, u=) = 0, u(x, 0) = Q(x) has at most one solution of class C1 in a neighborhood of x = 0.
Hint. This is Haar's uniqueness theorem of the given partial differential equation
(cf. the references in Exercise 1-14). Upon applying Exercise 1-14, with M = C = 0, to the difference ti(t, x) = ul (t, x) - U2 (t, x) of two solutions ul(t, x) and U2 (t, x), the proof follows immediately.
